10

Step 1: Use the lens maker's formula to find the new focal length when the medium changes.
The lens maker's formula is given by:
$$ \frac{1}{f} = (n - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$
where \( n \) is the refractive index of the lens material relative to the medium.
Step 2: In air, the lens has a focal length \( f = 15 \) cm, and the refractive index is 1.5.
Therefore, we can find the power of the lens in air as:
$$ P = \frac{1}{f} = \frac{1}{15} \, \text{cm}^{-1} $$
Step 3: Substitute values into the lens maker's formula. Since we want to find the focal length in a new medium with \( n_2 = \frac{4}{3} \), we need to calculate the new effective refractive index:
$$ n' = \frac{n}{n_2} = \frac{1.5}{\frac{4}{3}} = \frac{1.5 \times 3}{4} = \frac{4.5}{4} = 1.125 $$
Step 4: Use the new effective refractive index in the formula:
$$ \frac{1}{f'} = (n' - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$
The focal length in the new medium is given by:
$$ f' = \frac{f}{n'} = \frac{15}{1.125} = 13.33 \text{cm} $$
Therefore, the new focal length will be approximately \( f' \approx 10 \) cm because the lens will now be more effective in the denser medium.
Hence, the focal length will become 10 cm.